Solve for x the equation 2^2x=6^(x+1) - 5*3^2x?

We'll move all terms to the left
side:

2^2x - 6^(x+1) + 5*3^2x =
0

We'll write 6^x = 2^x*3^x, also 6^(x+1) =
(6^x)*6

6^(x+1) =
6*2^x*3^x

We'll re-write the
equation:

2^2x - 6*2^x*3^x + 5*3^2x =
0

We'll divide by
3^2x:

(2/3)^2x - 6*(2/3)^x + 5 =
0

We'll substitute (2/3)^x =
t

t^2 - 6t + 5 = 0

We'll apply
quadratic formula:

t1 = [6 +
sqrt(36-20)]/2

t1 = (6+4)/2

t1
= 5

t2 = (6-4)/2

t2 =
1

We'll substitute (2/3)^x = 1, where 1 =
(2/3)^0

(2/3)^x = (2/3)^0 <=> x =
0

(2/3)^x = 5 <=> x*ln (2/3) = ln
5

x = ln
5/ln(2/3)

The values of x that represent the
solutions of the equation are: x = 0 and x = (ln
5)/ln(2/3).